Question
Download Solution PDFWhat is the area (in unit squares) of the region enclosed by the graphs of the equations
2x – 3y + 6 = 0, 4x + y = 16 and y = 0?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
2x – 3y + 6 = 0
4x + y = 16
y = 0
Formula used:
Area of triangle =
Calculation:
2x – 3y + 6 = 0
⇒ 2x – 3y = -6 ....(1)
4x + y = 16 ....(2)
Multiplying equation (2) by 3
⇒ 12x + 3y = 48 ...(3)
Now,
Adding (1) and (3), we get
⇒ (2x + 12x) = (48 – 6)
⇒ 14x = 42
⇒ x = 3
Now, Putting the value of x in equation (1)
⇒ 2 × 3 – 3y = -6
⇒ 6 – 3y = -6
⇒ -3y = -1
⇒ y = (-1 – 3) = 4
So,(x1, y1) = (3, 4)
Now,
We put y = 0 in equation (2)
⇒ 4x + 0 = 16
⇒ 4x = 16
⇒ x = 4
So,(x2, y2) = (4, 0)
Again,
We put y = 0 in equation (1)
⇒ 2x – 0 = -6
⇒ 2x = -6
⇒ x = -3
So,(x3, y3) = (-3, 0)
Now,
Area of triangle =
⇒
⇒
⇒
⇒
⇒ 14 [Negative sign is always treated as positive in modulus]
∴ The required value is 14.
Alternate Method
When we find the vertices from the given equations
Given:
2x – 3y + 6 = 0
4x + y = 16
y = 0
Area of triangle =
Area of triangle =
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